Ta có: \(2\cdot cos\left(x-\frac{\pi}{3}\right)-\sqrt3=0\)
=>\(2\cdot cos\left(x-\frac{\pi}{3}\right)=\sqrt3\)
=>\(cos\left(x-\frac{\pi}{3}\right)=\frac{\sqrt3}{2}\)
=>\(\left[\begin{array}{l}x-\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\ x-\frac{\pi}{3}=-\frac{\pi}{6}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{6}+\frac{\pi}{3}+k2\pi=\frac{\pi}{2}+k2\pi\\ x=-\frac{\pi}{6}+\frac{\pi}{3}+k2\pi=\frac{\pi}{6}+k2\pi\end{array}\right.\)

